Solutions to PROBABILITY AND STATISTICS WITH APPLICATIONS. A Problem Solving Text. 1566987229, 9781566987226

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SOLUTIONS TO

~ PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

Leonard A. Asimow, Ph.D., ASA Mark M. Maxwell, Ph.D., ASA

ACTEX PUBLICATIONS, INC. WINSTED, CONNECTICUT

SOLUTIONS TO

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM

SOLVING

TEXT

Leonard A. Asimow, Ph.D., ASA Mark M. Maxwell, Ph.D., ASA

ACTEX PUBLICATIONS, INC. WINSTED, CONNECTICUT

Copyright © 2010, by ACTEX Publications, Inc.

All rights reserved. No portion of this book May be reproduced in any form or by and means Without the prior written permission of the Copyright owner.

Requests for permission should be addressed to ACTEX Publications PO Box 974 Winsted, CT 06098

Manufactured in the United States of America

109539) 65437221

ISBN:

978-1-56698-722-6

TABLE OF CONTENTS

Chapter 1: Combinatorial Probability 1 amet ie EL OD aD ymVIOGCle neha cck ystems Ser ook cceccs ac scrretnsch mites estes tcsscase 1 1.2 Finite Discrete Models with Equally Likely Outcomes ..........c.sccscccssssssssessssessesesessssesesees 2 eM ATU ara Ec UO Ut ON card stn sMeree cme era. oes. doses Sage ac acsTPe Letetseoes coset cus es eee 8 TES IN RGTOS TANbY)FCGE218(6)1S 1s a SR OH yah rR oe 10 meee DaDich IPS alliplcieXAIMINAtlOU, ten wet cemtese 5, tenth tctas ts ete roa sitec es eass aeece ee ee 13 Chapter 2: General Rules of Probability 15 PM CUBE THCOLV ASUL VIVAL Koll srtce sogeen toner ei ta coc ao Ateabevioscanavh cues Bem COUCH LON AL ELODADIILY 6-5 es ie ctseseen irra oo is Cie chetce ogi MR OCC CEC Ctra tn cet Nese ocuscaode Guat Cap yc taet a SNe SMES Ay CS eelNCOTSIYI Reet eia 0 ees, Ws each Ty Re nD ee SN Re in occy0. cen Lia CoECOMDUINCY tee Pewee Ree PME Pee IU cles aliple EX aNMDatlOl sc sc1 tae tg a ers AR Aca,

(Eee vues: we Resa cea eee cree AC Eon cea oa CMR aN oe ee AS NCO ako e a Ue eed sea ee ee eee rye hioacts anes eserth ioe ene:

15 23 29 35 38 40

Chapter 3: Discrete Random Variables 51 EMIS LeosANCOMI.Y ALA DICS omen eet es aa aiced Sithcee artiste ors ete ats Pee ULALIVe EroUaDIlLy LiSitiDULIONe er a etc eee a ee ca eres ne Boe Mc cotirC ayn (ci ralel CLICCIICY aerrets eee nicer: ars ahans ys OC eae mer ciaeei od tos Eee Re HC ASUITOSM aLISS CE SILOM erat eotner te tommeted cseeraten ty tnReocia aaaraat eek ante een cre ee Pen ONKOL ex DeCLAUON ANG. V ALANCO. sara sees eoceaceed.coasece cassdndsaphpeueetepbes nian eset

51 a2 a 56 ao

SomemrOiny listed

Random. Variables (ROUMnG 1) i ..jcc-+..tss avn cavesoecasncssouevasttocweansndeones 62

Soe EXO UA IMI yA Oe LAtIN ee:PUNCLION: sxc.nsatectscsseet sides secs(oasis siieutcnseiariensnvacnnsuaancas@nesnsomes 65 Fea 11) teeny MUNA KATIE Oar cee ocean stent ovicaog ict oie wn sieancano dnees dun ee vacua soutemen neue haeak oe 66 Chapter 4: Some Discrete Distribution 73 SP Uist emTOUR LUE UNIOU wetge sscles rncger eeniveneeie ate ar sha ce dshy vara Snnca>yeruacieseee 73 aoe Deraoill triale and the DYimormial DISPUTOD cose tosscxencsetievanceraesesseusumagesutens-cosacanas--nadees 76 SPM rere OIAC ITTGR!ISELIN eee ecg regi can 5 Coxe ascent e datspsoesdupiaa seep asuacgdth da aalia soto ene danon 78 em ee ASV ESOT JISCUIU LOU oponcsscacarnc gapeaanissmny smcpig-te ths wwaantaasypGrtaicduoasneduda scien sve’ 82 Gee CAELY et eCICOMICITIC. IST DULIOM so,ucct:csocoecrs sndeuse sui oorensaftrsantpach coatnens saataacaruruspomcanan' 83 Sm LICH ASSET SSL PUL LULU ety orecar rie aetraks avacoedvecout creck y ck cata tanta ri dhaain sadlsea is nstuas haa hinn 84 vinesinestnsuserensny areananennrey nan are89 GIALS CoOUDArisoll OL LISCLETS LISI DULONS divyorenccenss. BES c Napict 4 matinic EX AIMAtION eerste css enshcdtns.svoegy re costal atactraecanaban< argon: 93 imme Chapter 5: Calculus, Probability and Continuous Distributions 97 LLOM ae tiasciiaa rant cater cb es? IV Assaraly as cof wats dunes soon nt vusch,casiornas iN adbssacnunanaesa ise. oF me ML eS IEE UMC MLC erPCC ON Goat ot crepe cimn ate cicwat se by sillnish oo dieovsands inges nyunsoaatenges capsinnia svsis>oxslenes quantum cannes: ao SOCEM COLL VESEY ASUONLstpee Cee te usta RY ol cae sav capaha unsaaautin ics vbrannntoactyts genk oc sSaencns sens104 Applications to Insurance: Deductibles and Caps.....cccicesecseseetnessssessnessessseeseesenens 105 5.5 fetapencsanacsien euapnqects og nen 107 esisatsreats enncaacegiqayunts Oe tCC ETAL PUG CULO Mes ecaeesagtevac UV sesenntsasnsoneens 111 sssentosestesseeaeesseneoeno scisscisesccscsstsesacossces Chapter S Sample Examination... 5-7

il ® TABLE OF CONTENTS

Chapter 6: Some Continuous Distributions 115 C7 ee UDLLORTR ANG OMIeV ALLA IES acccas seater, ete nee eae ees Py eee tae pee giant tec 1 Ore) Hee XPOUCA LDIS(T DULION tc... toes xt nee aes em aioe oe yc ah lk och eneaete abe 116 6:3 [he Normal Distribution... at-0 ccm hese epee wood ocd ceo encstou ocpinsonnjaaenetee= ce 120 64 Pe ihe law or Averages and the Central Lrmit VR6 Gren ies. scstecesss-cee-ccdocesesoacvaraneudeevatene122 Orme Ne GraTniina-L)ISCITOUILL OIL cp tere crete tne ee eerede- fag ete gine te dnc ven ats atee saa nie altuna eevee | Wee G Geel Ne-Beta tainty Of L1Sttt DUNOMS se eecna cess oavevees teeter er tor et ws eaunee aa ore reese wader 127 Ow Ole © ONUTUOUS LIIStPIOULIOIS sree eastern tate neti gtentees nt tesa ea asian tacoe cies 130 GrSaar CMaptcy O-Salaple EX ALMITALIOL sa corse cr sncee ieee raed aster eer eee meine adieree tgs ecteoacernee ened 132 Chapter 7: Multivariate Distributions 135 7A, J Olt Disiibutionstor Discrete RaiCOM VAMADIES jcessesee pevercses-cercccet eecnecn-auneekeeteneeene 135 7.2. Conditional Distributions — The Discreté Case... iesc-c-secenscnen--ee ian aaitaces dome ee 137 Jedme. Independence. DISCTELe, CASE ares teccasiynnt neta. ee URC ee aati alse e ee epee a 139 eee COVarIance atid COTTCLAUOU:.«.a5 thet eoeaccts sascha cee erence Se nace tat clas eerie a ease 139 7 yee Joint DIstioutions for Continuous Random Wallabies .rec..sstece-e ee asset eee 143 HeOme SOUCIHONAL WIStUIDUM ONS =. He G OMINOUS ese peer ens see css nase aaa 147 Ueyeeincependence and, Covariance m thie: CONUMUCUS, CASE. 5, ..cc.cncesersseten coestnaesesendesnca tanec 149 Pp mee Gy armale NOrmal: LISI DU TOL scum see coe tccve: erreueeaeroe tor ae ee deseasodedia: eked tenn ne eenenenmetetee 154 fal Cevioment Getlerating Function fOr a J OUIt DISH DULLON ys. -c oop aceeeveonanryoe ee-cneacento 155 POMS AO Le igieS ATL EX AIIIITALION sernorce a hleOocorenetacus tien temsuna ant ase take eased aman a ee 155 Chapter 8: A Probability Potpourri 163 Sree be istlibution Obdslranstorimed. andor: V alialee gece. ates ee orem eden een Onl meetHe WIGMENE Gener atlily PAIR CU Oi ser gecsuce au. oc steeee te Meron cou eee ete ees oe nee S cig, (COV anIaNCe! b OLAlaser sam, meee paces ec sctreenscrcee ce an easuanuecs teeneaeeee ee ee DP sa ohn SAge he Coma iionin eek Oni las eerste. soe beeen cae nec ae orare oe eee ee Se CHAplEr GS Saliple BXaMie AOD. asc cceececs ce tete crvaeiecesase rene roeee races mace ein eae Chapter 9: Statistical Distributions and Estimation 187 Dales bhessample: Mean asran PStimiatot rye ccc ctor te aecree ee eee eee

De eeu

Demerol

Mating Lee OPUlAatiOn. V alLALICOs,

ic StNGent 1 Listhi DULL

cxccac ecco es er Glatt

eee

a eee

rercsri tarts ca ieennes © oanrsts Uneaten teem

eo certo tao

163 178 179 179 183

187 ae 188

ee eens bone AS 190

De Mme eetLEEDS LAD ULL Lyn teste’ cee tvs sua cel cares astieaeoutenen aueeie rene ter osseuen vceanvee ea tee eames aeneremete eRe SP am SUITIATILLS PTODOLLIOLS 5 .acace-cesoncacovsaty oxseveaeu roan Uatetens teeta aataunreacapsr ae sar tee eae eee tees DEO MeL stiiiaung thea iierence, DOtWweenl IVCALS seed sectre teeet ens enisstecsteeeat waenateemetee ea treme Fee SLULENAGIIVG CLS eSALII()LS cOUZ eattgmn nrg ei eyes uate seeks oneniert ceemesits cactiacercere caeta tact eter ee anne DS COADLEr DO Sar pLe Haima COM. cx csncngsnseit eines teas tsaneenceees auevenie aan santesesgesccsnte nena enema

191 193 194 195 196

Chapter 10: Hypothesis Testing 201 LOST Hypothesis: Vesting: Prarie work: nasa shen canchs sce seuraaite nine arth cache ttee tek eat ee en 201 10.2 Hypothesis ‘Testing for POpUlation VICeIS cencconreuseanreenastcmreeectcs¢eteeentnesee eee nena 201 10.3 Hypothesis T estings Tor: Pop UlariOne Vartan a acccecetenigue terrence unten otters eee ene 10.4 Hypothesis Testing Tor PropGmOnSwan. anexs-c 10C1 +5 Cy + 20Cy > 25Cy > ani «10 C2 > 5 Cy + 99C) + 25Cy + an Cy - 19 Ch - sC2

= 47,500,000

ways.

47,500,000

bird ofof each each variety) £192,052. 400 = .033 985. Pr(atr(at least leastone bird variety) =~

ee

Pr(passing on first attempt)= 1oS)

Use the multiplication rule with the following steps: Step 1. Step 2. Step 3.

Step 4.

Choose List the Choose Choose Choose

any 3 rows (order immaterial) - # ways = 6C; =20. rows in ascending order for definiteness. a column for the first listed row -# ways =5. a column for the 2™ listed row -# ways =4. a column for the 3" listed row - # ways =3.

Answer = 20x5x4x3

= 1200

(C)

CHAPTER 2: GENERAL RULES OF PROBABILITY Section 2.2 Set Theory Survival Kit

2-1

(a) (b) (Cc)

(d)

O=rl3 od.7-9} and, Pat 2 aio. The odd primes less than 10 are OM P = {3,5,7}. The set of odd numbers or prime numbers less than 10 is OMT Pee 28 5.7.9. The odd numbers less than 10 that are not prime is the set

Age

(a) (b)

B ={az water}, “so N(B) =2.

(c)

AUB=(1,2,z,Jamaal,gum},

(a)

The Luisi family pets can be summarized as:

(b)

The number of pets of each type is summarized as:

so (AU B)’ = {water}.

Named

\ O— P = {1,9 Sf:

16 ® SOLUTIONS TO EXERCISES — CHAPTER 2

2-4

The shaded section denotes the indicated set.

2-5

From Subsection 1.4.3, we already have the probability of the winning prizes. Pr(losing) = 1 — Pr( winning something)

> Rye .

le

ae

1,712,304

120,576,770

120,576,770

probability of winning the grand prize

probabilty of matching only the powerball

_ 117,184,724 120,576,770 °

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 17

Doubles = {(1,1),(2,2),(3,3), (4, 4),(5,5),(6,6)}. Sum is divisible by 3 = {(1,2),(1,5),(2,1),(2,4),(3,3),(3,6), (4,2),(4,5),(5,1),(5,4),(6,3),(6,6)}.

Doubles - Sum is divisible by 3 = {(3,3),(6,6)} Doubles U Sum is divisible by 3 = {(1,1),(1,2),(1,5),(2,1), (2,2), (2,4), (3,3), (3,6), 1)

a)

0,1), (024), (,.2),

(0;3),.020)}

Pr(Doubles U Sum is divisible by 3) = + = < + oa— =

Let K equal the percent of king size mattresses sold, O equal the percent of queen

size mattresses sold, and 7 equal the percent of twin size mattresses sold. We are given: K+Q+T Solving for the variables,

= 100%, K

Q.= LK +7), ane Kae

3 2

=60%, Q=20%, and T = 20%.

The probability that the next mattress sold is king or queen size is K

2-8

A’ = {2,4,6,8,10} B' = {1,4,6,8,9,10} AUB ={1,2,3,5,7,9} ANB =33,5,7) A'UB' =(1,2,4,6,8,9,10} A' 0 B' ={4,6,8,10}

(AUB) ={4,6,8,10} 2-9

A-B = ACB’

is represented by the shaded area. A

B

+Q = 80% (C).

18

D210.

SOLUTIONS TO EXERCISES —- CHAPTER 2

“Cte eee

(4UB) =4'0B'

2-11

AU(BOC)=(AUB)A(AUC)

2-12

N(AUB) = N(A\B)+N(AQB)+N(B\A) [Nv(A\ B) + N( AMB)|++[ N(B\ 4)+N(ANB)|-N(AOB)

= N(A)+N(B)—-N(AMB)

_ PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 19

2-15

(4 Bo

c) is the shaded area

AM(BUC)".

2-14

Method 1: Using the Axioms

N(A) N(B)

6 = N(U)-N(A). 10 = N(U)—N(B’).

From the inclusion-exclusion relationship,

N(AO

N(4U B)+N(AMB)

B)=6+4+10-12=4.

Method 2: Venn Diagram

Method 3: Venn Box diagram (Subsection 2.2.4)

= N(A)+ N(B).

20 @ SOLUTIONS TO EXERCISES — CHAPTER 2

2-16

NM(AUBUCUD)

= N(A)+N(B)+N(C)+N(D) —-N(ANB)-N(ANC) —-N(AND)-N(BOC) —-N(BAD)-N(CAD) +N(ANBOC)+N(ANBOD) +N(ANCND)+N(BOCOD)

—-N(AQNBOACOD)

There are 4C, =4 ways to select one of the four regions. There are 4Cz =6

intersections of two of the four regions.

There are 4C3 =4

intersections of three of the four regions.

.

2-17

A four region Venn diagram could be drawn as:

2-18

There are seven male children without purple hair in this family.

Adult

Female

Bin Purple hair

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

2-19

(B) 50% = 40%+10%

of the population owns exactly one of auto and home.

Auto_owner

2-20

(D)_

Home

There are 880 young Tharried females.

Married

2-21

(D) 52% of these viewers watch none of the three sports.

Gymnastics

aN

yy Soccer

Baseball

@ 21

22

2-22

SOLUTIONS TO EXERCISES —-CHAPTER 2

(A)

5% of the patients need both lab work and are referred to a specialist. The percentages in bold are given in the problem. Try this just using a Venn diagram. Lab work | No lab work Referred to Specialist | 5% 0 Bier ae Not Referred to Specialist 35%

60% 2-23

(D) Pr(A) =.60. The key here is to note that Pr(AU B) = 0.7 implies that Pr(4’

B') = 0.3.

Similarly, Pr(A U Be)=0.9 implies Pr(4'n B) =0.1. These are versions of De Morgan’s laws. We do not have enough information to find Pr(A A B) or

Pr(AB'), but fortunately we are just asked to find Pr( A) = 1—(.1+.3) = .6 A

(D)

B

We have two equations with two unknown variables x and y (see box

diagram below). The first comes from the fact that the sum of the probability of four disjoint outcomes must equal 1. The second equation comes from the statement of the problem. 22%+x+y+12%

22%+y

= 22%+x+

—.,—_——“~-

We

probabilty visit chiropractor

x = 26%

= 100%

14%

x+y

= 66%.

imphlesthat

y = x+14%.

—-—-

probabilty visit PT.

and

implies that

exceeds by 14%

y=40%.

Pr(visit P.T.)

= 22%+26%

= 48%.

Chiropractor

22% + y

No chiropractor

x+12%

22% + x

100%

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 23

2-25

Draw the Mickey Mouse diagram, with Auto, Homeowners and Renters labeled as

shown below. Since H and R are mutually exclusive the subsets comprising their

intersection have cardinality zero. Statement (i) gives N [AUBU cy |Sa, Statement (v) gives N(H \ A)=11. The remaining unknown subsets are labeled

eye Z AWN

From (ii) x+ y+z = 64. Also, 100 = 174+114+(«+y+z)+w From (111), y+11

From (iv), x+y gives

Section 2.3

2-26

= 174+11+64+w

=>

we

8.

= 2(x+8), and,

= 35. Solving the last two equations simultaneously

x = N(AXMR)=10

(B)

Conditional Probability

There are 3 levels of choices in the following tree. The first branch concerns whether the professor put the batteries in an empty pocket or in the pocket with the two new batteries. The second level concerns which pocket the professor withdraws batteries from.

In the third level, if the four batteries are in the same

pocket, we will use

counting techniques from chapter | to determine the likelihood of grabbing either two batteries that do not work, two batteries that do work, or one battery that works along with one battery that does not work.

24

SOLUTIONS TO EXERCISES

— CHAPTER 2

Professor selects

Professor places dead batteries in empty

;

Select two Bee .

1

pocket with dead

eee

batteries

:

3 we : Professor

pocket

er

Select two

selects pocket

1 __ good batteries

with good batteries

2

‘alt

Select two dead batteries

Professor places dead batteries in pocket

1

with 2 good batteries.

4

oe

| gi ees l Fe

1

4

elect one

professor selects beket with all :attenes

dea

ie

and one good battery Select two good batteries

(a) (b)

It Pr(get at least one good battery) = =.Bers

1 ») eee eee

Peay

G

Pr(two good batteries at least one good battery) _

Pr(two good vat least one good battery) Pr(at least one good battery)

2

Pr(two good)

Pr(at least one good battery) ;

mall

Tiga,

ae

4

24

(c)

l able l

play

m0)

4

Pr(batteries in same pocket|at least one good battery) Pr(batteries in same pocket Mat least one good battery) Pr(at least one good battery)

The audio device will play

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 25

2-27

(a)

This is the scenario where we will keep our original door. You will win the car if and only if you selected the correct door originally (it just doesn’t matter what Monte does). There is a one in 3 chance of this. Pr(win auto|keep original door) = +.

(b)

If you originally select the door with the new car, then Monte can select from either of two doors to show you a goat. If you select a door concealing a goat, Monte only has one door that he can open. Consider the following diagram. shows you

Monte

:

You pick a door

hiding a

If you change doors, ?

ei

the goat below

l

you win a new car

goat ou pick a door hiding a

Monte shows you aA

goat

the goat above

If you change doors, >

|

you win a new car

'

3

If you change doors,

Monte shows you

a door hiding

the first goat

you win a goat

vof—

shows you the second goat

Pr(win auto change doors) =

~

i)[o High] = Pr[High|[rregular]- Pr[Irregular] = +-15% — hi a = Pr{Irregular

Similarly, from(v),

c = +: 64% = 8%.

Regular Irregular

High | Low | Normal ee

Row Total

oy

SES ies

85

5

2

8

15 100

Column Total

Pr[Regular ~ Low] = 20%

Completing the Box Diagram gives,

(E)

|

26 @ SOLUTIONS TO EXERCISES

2-29

- CHAPTER 2

(C) Let x equal the probability that the male is a smoker, given that he does NOT have a circulation problem.

Pr(circulation problem] smoker) =

Pr(circulation problem ~ smoker) Pr(smoker)

oe ak

5

5 Oe

ee

4

ae

Smoker

2x

Smoker

x

25%

circulation

problem

2-30

Cyne

(C)_

2-31

(B)

Pe

oS

ees

: Pr(smoker A died) .10-.05 .005 ker|died) = ————-___—. = —_——____—_____ = —— _ = 36°

Pr(smoker/died)

Ne

Pr(died)

) Pr(Heads-Heads) = (2) ae, 3

2

.

3

Head

Tails

ae

Pr(Heads-Tails) = eae

2

Px(Tails-Heads) = 44 : =

3

é paresip

3

Heads

Tails l 3

First coin flip

|

Maes

pyTails-Tails) = (+) = 3)

Second coin flip

Number of Heads

| 0 | l

Probability

9

cathe thoagh the srobsbibiy

tree that have one head and one tail

384

tee Make

Makes his 3 point attempt

384

384

Miss

616

Nree

Misses his 3 point attempt

Pee

1

Make

: 616

Make

384

Miss

616

Make

384

eit.

616

384

John Stockton’s

_

Miss

616

Make

384

Miss

616

616 Ea

first attempt second attempt

third attempt

Pr(Stockton makes exactly two 3-pointer) 3

384

.384

.616

ways to make exactly two 3-pointer on three attepts

probability of making a 3-pointer

probability of making a 3-pointer

probability of missing a 3-pointer

= 2725

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 31

2-40

iy Pr(B| A)=—.

2-41

(a) Since Pr(B)=b, we know Pr(B’) = 1—b. To show that4 and B’ are independent, we observe Pr(4> B’) = Pr(A)-Pr(B’) = a-(1-b).

Pr(B) api

51

The events A and B are dependent.

52

Q

Ca [a RRS w=) 2-42

2-43

B

B'

b

eoah)

‘Not necessarily.

(a) (b) C = {HHH, HTT,THH,TTT}. Thus, Pr(A) = Pr(B) = Pr(C) = 4. (c) Pr(AB)=Pr(HHH,TTT)=2-=1.

Pr[A

2-44

Similarly,

VC] =Pr[B AC] =Pr[HHH, TTT) =.

(d)

Yes, because for example, Pr(AM8B) = + = +.

(e)

Pr(4

(f)

ING Weiter Arie

Let

x=Pr(AM 8B). Using independence, solve (.2+ x)-(.3+x)

SGIUUONS, X=

=ePTCA)- PTC):

BOC) = Pr[HHH,TTT]=+. Cyt

ee

= SPs

A] PLB) Pr.

= x. There are two

2 0lL =.

Pi

eet Obar 0.

Since the coin is fair (all outcomes are equally likely) and we do not need to track probabilities, listing the sample space will be sufficient.

2-45

U =|HHH,

HHT, HTH,HTT,THH

A = first coin 1s tails ={THA.THT

B =the third flip is heads =

THT ,TTH,TTT. TTH,TTT}.

{WHH, HTH ,THH ,TTH }.

(b) Yes, the events are independent.

32 @ SOLUTIONS TO EXERCISES — CHAPTER 2

2-46

(a)

Now is when tracking probabilities through a tree diagram is critical.

agPr(B| A)a (c)

2-47

PrANB) =_ 4-.6-.6+.4-.4-.6 = ———.

Era)

= 60%.

A

Yes, the events are independent. You can see this by computing Pr(B) = Pr(third flip is heads) =.60.

Let O be the event of using orthodontic work, F fillings and E extractions. Using (1) and (11) we have:

tele

eae

Ss

Gerth

Ol) Primal

1—21o—9

b=j12=1)2. By independence, Completing the O and F box gives:

3

d= be=>c=a/b=213-

Row Total

1/3

Bes)

O | O' | Row Total

|

E

1/4}1/4

le2

E'

[1/4 ]i/4]

1/2

|Column Total | 1/2 | 1/2

We now use Pr| F |= ly

|

|

anc Pri Efe l /2, plus (iv) to fill out the F and E box:

ip E

[

F 1/8

E'

Row Total

LS | al

Li

7/24

1/2

Column Tota!

Note: F and £ are not independent.

Pr| EU F )=1-Pr| EV F'|=1-7/24=17/24 (D)

l

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

2-48

33

We take the statement that a “different brand of pregnancy test was purchased” to imply that these test results are independent. Pr(both tests are incorrect) =.01-.02 = 0.0002.

2-49

First we fill in the missing information.

ivajor | ee Grades yeas | TAT BIC] DE |Totals | | Business| 2| 3 |3 [5 [1] 14 | PSs lem ee | Totals |9 |10}4[/5[3] 31 | (a)

Pr(student assigned ‘B’) = =. students - business majors

: b (0)

: : number of science majors — 17 Pr(science major) = -———_-_—__>—_ = — = \ yor) number of students on

(c)

Pr(science major ~ assigned ‘B’) = +.

(d)

This is most efficiently calculated by looking at the table. We are told that the student is a business major. Therefore the student lives in the shaded row of the table. Pr(assigned *C’ b usiness

2-50

major)

31-14 at

3 et is 1

(e)

Pr(business major assigned 'A') = =.

(f)

No, because

(a)

Wewill use counting techniques (combinations) that we studied in chapter 1. The number of ways to select 2 batteries from a flashlight containing 5

Pr(assigned *C’ business major) # Pr(assigned *C’).

batteries is 5C> =10.

Pr(both batteries lost their charge) = at eho:

(b)

Pr(exactly one battery lost its charge) =

:

=

(c)

Draw a tree diagram with appropriate probabilities. >

Pr(exactly one tested battery lost its charge) = 2-

bo in|

34

2-51

SOLUTIONS TO EXERCISES—~CHAPTER2

~

Let Pr(purchase disability coverage) = P[D]=.x, then Pr(purchase collision coverage) = Pr[C’] = 2x.

0.15=2x? implies that x =.274 and 2x =.548.

The completed Box Diagram is:

ee KE oa [ea Sonn Tour]sasa2[1

Pr(select neither insurance) = Pr aoN D'] = 328.

2-52

(B)

Pr(select two consecutive face cards) = =-- ae or using combinations Pr(select two consecutive face cards) number of ways to deal 2 face cards

12C2

number of ways to deal 2 cards

2-53

2-54

52C2~

(a)

Pr(next roulette spin is black) = 4.

(b)

8 Pr(next eight roulette spins are black)= (48 = U02 53:

(E) Pr(no severe and at most one moderate)

= (0.5)? +0.5-0.4+0.4-0.5 =.65.

minor

moderate

severe

0.5

0.4

First Accident

Second Accident”!

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

2-55

(A) Let B denote the number of blue balls in urn 2. Pr(both ball same color) =.44 = (0.4)-

16

16+B

B

+ (0.6):

_ 6.4+.6B

[6% Bael6e

Solve for B to find there are four blue balls in urn 2. 16 16+8B

16

16+B

a4 10

Red ball

B Waa

Blue ball

16 Red ball Blue ball

16+8B

2

ae

10

B

rm

Blue ball

16+B

Urn 2

Section 2.5

2-56

Bayes’ Theorem

Draw the standard tree diagram to track probabilities.

~ 51 Ace

pais on

52

3 2 Non-Ace

51

Sade Non-Ace

48 uy

51

First card Non-Ace

A 51

Second card

:

number of aces

-

Pr(first card is an ace) =

(b)

Pr(second card is an ace}first card not ace) =

number ofcards

4

a

(a)

52

35

36 @ SOLUTIONS TO EXERCISES —CHAPTER 2

(c) Pr(first card is an ace|second card not ace) Pr(first card is an ace ~ second card not ace) Pr(second card not ace)

Be

°

AGS

AS

ip

Se

FAT Sil

2-57

40

11Face

50

a

40

eh See 5D

Non-Face

Face

sy) First card

Non-Face

Face

11

Ae

50 oh

51

on-Face Non-F

39 50

Face

50

51

Non-Face

=

39

Face rr

12

= Second card

&

Oo

50

12

ae

5

Non-Face

Third card

Number of Face Cards

Probability

.F¢

A

0 9,880 22,100

9,360 22,100

2 2,640 22,100

+ 4

220 22,100

2,860

(b)

Pr(at least 2 face cards) = F100"

(c)

Pr(3" card is face|first two cards are face cards) = Pr(all three face cards |first two cards are face cards) = e

PROBABILITY AND STATISTICS WITH APPLICATIONS:

(d)

A PROBLEM SOLVING TEXT

# 37

Pr(all three face cards| at least two face cards)

___Pr(allthree face cards)

220 — 22,100 _

220

2,860

2,860°

~ Pr(at least two face cards)

22,100

(e)

Solution Method |: Tree diagram and Bayes’ Theorem Pr(at least 2 face cards| last card has face) a

Pr(at least 2 face cards ~ last card has face) Pr(last card has face) this is the probability that the first and third cards have faces

.

iP Mim = 1 0Nt B40 gio 5ST 50 Bes 50) Oh oD Py i Pane ol Th eee 3) GS i i SS ee = This equals S Does it make sense to you?

— 11880 _ 3999 30600

Solution Method 2: Using Combinations

Imagine that you know that the third card is a face card. That leaves 11 face cards in a deck of 51 total cards. Now imagine dealing the first two cards from this reduced deck. The question asks us to find the probability that at least one of these cards is a face card.

40 C2 Pr(at least one face) = 1— Pr(two non-face cards) = 1— NE se Hey Ge 5102

38 @ SOLUTIONS TO EXERCISES— CHAPTER 2

Section 2.6

Credibility

2-58

Ay

0.10

Good driver

Bad driver

Type of driver

A

0.50

Accidentor not in year| Accident or not in year2

(a) PrA. As) =.8201)* 2-5)?

Cha

=.058.

Pr( 42 OA SoG Lyhed) SA bee pete mee Pr(4,)

(c) Pr(G|A ro yy

28.5 (1) ba2ed)

ese

ater

Pr( A, C

SS

boa

29

>. (1a. 2 < (d) Pr( BlAt least one accident) = ae Lee aaa .2-(1-.5*)+.8-(1-.9°) 302

497.

PG

A Az) As A

Pr(A, 7 A>)

2-9

a)

2-60

(D)

mel. )s + .2-(.5)-

008 imi O58

Ales: (e) PROTAro AAS

A>)

O58 LEE 18 90

8 .(.9)2 he)

_

.8°(9)*4+.2:(5)e

Pr(1997|1997 or 1998 or 1999) = =e .16-

_ .6482 — 9284.

~ .698

ae

(.05) +.18-(.02) + .20-(.03 )

= 455.

Draw atree diagram.

Pr(ultra-pre ferred dies) =

(0.10) -(.001) =.0141. (0.10) -(.001) + (0.40) - (.005) + (0.50) -(.01) a

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 39

2-61

(D)

Drawa

tree diagram.

(.16)(.08)

Pr( young adult collision) = (.08)(.15) +(.16)(.08) + (.45)(.04) +(.31)(.05) —

2-62

(D)

Let xdenote the probability that a non-smoker dies. Dies

Heavy

20%

Smoker

|. Lives

1—4x

Dies

Smoker

4x

2x

30%

Lives 1-—2x

NonSmoker

50% Type of smoker l-x Live or Die?

fi

fo ce

Pr( heavy smoker}died) = (.2) +(.3) (4 (2x) x) +(.5)x

lp 19

(.08)(.06)

it)1 lon)ios)

(B)

= Pr(16—20jaccident) = (.08)(.06) + (.15)(.03) + (.49)(.02) + (.28)(.04)

158

40

SOLUTIONS TO EXERCISES

Section 2.7

- CHAPTER 2

Chapter 2 Sample Examination

1. A

Z.

B

Pr(dealt a fullh NE

Cus)

eeber

of full houses ee number of 5 card hands

me 13Ci C519

Ci aC a

3744

52Cs

3.

2598960 ©

You may wish to create a Venn diagram. The key is to note that that you must have some insurance to be a client.

Auto Insurance No Auto Insurance

Dismemberment | No Insurance Dismemberment il7/ 45

(a) N(both auto and dismemberment) =17.

(b) N(exactly one kind of insurance) = 65 = 2(

(c) No, because Pr(auto ™

dismember)=

17

a aw NM

62

go * 95" 89

= Pr(auto) - Pr(dismember).

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

@ 41

(a) Venn Diagram

U =College class

Engineers

Female

Venn Box Diagram

(b) Pr(female > engineer) = a oo (c) Pr(femaleU engineer) = ca = pit

a

a2

by)

(d) Pr(female| non-engineer) = =. (e) No, since Pr(male Mengineer)

: Ke

ea

aey)

# Pr(male)- Pr(engineer).

number of ways to be dealt 3 Kings =

:

=

number of ways to be dealt 3 cards ES

(b) Pr( NOT dealt 3 Kings) = 1

Cy

ce

Oy

22100" 9)

(c) Pr(3™ card King|I‘' and 2" card King) = av

(d) Pr(3 Kingsat least 2 Kings) =

4C3 ae ancy

=ACTON

=

4C3 Z

52 C3

=

4

22100

e

42

6.

SOLUTIONS TO EXERCISES — CHAPTER A)

(a) There are 16 elements in the set.: The bold numbers denote a sum of 6.

‘a= {22,23,24,25,32,33,34,35,42,43,44, 45, 52,53, 54,55}. (b) Pr(sum is nine) = =.

(c) Pr(at least one die is 5) = (d) Pr(sum is 6at least one 5) = (ec) Pr(at least one 5|sum is 6) =

It

ws, 9° we 3

(a)

(b) Pr(sum is 4) = (c) Pr(at least one 2) = (d) Pr(sum is 4) ball from Urn 4 is 2)= Pr(ball from Urn B is 2) =

(e) Pr(ball from Urn B is

2/sum is 4) =

~

=

i st

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 43

8.

Se] Coke

i |Ripple | Water |Red Bull|__|

eager [alte dt oaMeeSESEY eee [aes |e facia lela Ma cl PE arian SO Oe ese es Se 0!

|

b) (b)

5 Pr(Red Pr(Red Bull) Bull) =—30

(c) Pr(female| milk) = N(female

4 milk) 8 Namilk) SG

d) Pr(male U milk) = 22. 307

(a) Pr) =07)

(b) Pr(A’)=0.6. (Cy PH Arm.) =.255

(een eyPr(AA B) Pr(B)

(eyo

Pr AGB

ORCI

)\=HOs.

Pr(B es eerra A) _

(g) No, they are dependent.

‘40.

Pr(4 7 B) # Pr(A)- Pr(B).

10. Assume for definiteness that card with 5 credits is in the left (L) pocket and the 4 credit card is in the right (R) pocket. She could either spend all 5 credits in Z, or spend one from L and 4 from R, with the last one coming from R. These actions can be displayed as the following 5 letter words using LZ and R: LLLLL, LRRRR, RLRRR, RRLRR, RRRLR. 5 l 5 Each word has probability Gy , and so the answer 1s 5- +) ‘= 32 5

44

RL.

SOLUTIONS TO EXERCISES

—CHAPTER 2

Assume first that the left pocket is depleted leaving 3 credits in the right pocket. This means a total of 6 credits are spent, | from the right pocket, 5 from left, and the

last one used is from the left pocket. This can be portrayed using 6 letter words with 1 Rand 5 L’s, ending in L. That means the first 5 positions in the word consist of 1 R and 4 L’s. There are

5;C; =5 such words.

Next, assume that the right pocket is depleted leaving 3 credits in the left pocket. This means we have 6 letter words ending in R with 2 L’s and 3 R’s in the first 5 positions. There are sCz =10 such words.

122

Thus, the total probability is

Consider what can happen to the team that is leading 3 games to |.

50%

Series won 4-1

Win

50%

Series won 4-2

Win Lose

50%

50%

Series won 4-3

Lose Game 5

50% Game 6, if

necessary

Lose 50% Game 7, if

necessary Pr(win) =1—.5° =.875.

13;

Pr(win) =1—.43 =.936.

Series lost 3-4

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 45

14.

Column Total

b=1-1/3=2/3 and a=Pr| AN B|=Pr| 4] B|Pr| B]=

pes

Tee

Lk Complete

the Box Diagram to give: Row Total B'

eS

Column Total | 2/3

Beal

13

y

1

Then’ Pra B)=1—2715=13 715. tS.

Begin by writing out the sample space of possible dance couples. Betty’s

Danielle’s

Dance Partner

Dance Partner

Danielle Danielle

Dan | Boris

Dan | Danielle | Andy Andy | Danielle | Boris

Pr(3 wives dance with non-spouse) = 2.

U = fabed,abdc,achd,acdb, adbc,adch, bacd, badc, bead, beda, bdac, bdca cabd,cadb, chad ,chda, cdab, cdba, dabc, dach, dbac, dbca,dcab,dcba}.

Pr(all wives dances with non-spouse) =

Nae tO

46 @ SOLUTIONS TO EXERCISES

We

-CHAPTER 2

There is one good young teacher. U = math faculty at School University

Old

18.

Bad

Pr(first person wins)

Wei

Sea

Nae

‘On

5

Wy

AY

5 Red bal

i

Se 4

Red bal

White

ball

4

The

First ball selected

=|~

Red bal

=

White ball

IZ

Second ball selected, if necessary.

sak, l

White ball

Third ball selected,

if necessary.

19.

(a)

i

2}

Pr(1* person wins) = Bama +| 6

6

7 -—+,..=40%

¢5

Use geometric series.

(Oy

> + ers person wins) =. G26

L978:

eos CT COeTera

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT ® 47

20.

To solve for x, we know:

a =Pr(A44 BAC|ANB)= 3 Pr(no risk factor| A)= =~ =.46.

Zzby,

PHA

BAC)

= ex

Prodan yee ao ee

(C)

(C) We have four equations with four unknown variables: 1 5 it : a , and w+x+y+z=l. ee 4

Vg

et

e

12

e

Adding the first three equations together, we see 2D 22 =

=>

Xxytzy

= |nN —

Pr(no coverage)

=

w=1-(x+ y+z)=1 = = 5

0G

48 @ SOLUTIONS TO EXERCISES —-CHAPTER 2

De.

Let A represent the event of heart disease. Let B represent the event of at least one parent with heart disease. Then:

Pr( 4]B’) = —_ = 1712.

ZS,

(B)

P| One Car |Multiple Cars | Row Total | | SPOnICACIeNs |ae St en ey NSIS ports Cates |Seeea | Stae ee eee Column Total a

70

Pr |Sports Car M Multiple Cars | Pe |Sports Car |Multiple Cars |Pr |Multiple Cars |

(15%)(70%) = 10.5%. The completed Box Diagram 1s:

pS One Car |Multiple Cars |Row Total | Sports Car 9.5 No Sports Car | 20.5

Pr(one car ~ no sports car) = 20.5%

24.

(B)

The Auto/Homeowner Box Diagram ts:

Thus, 35% have auto insurance only, 50% both. Then Pr(renew)=

(.50) + auto only

(.40) renew given

insure only auto

have homeowners only and 15%

~=+(.35)-(.60)+(.15)-(.80)

= 53%.

(D)

have

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 49

25%

Begin with the Emergency/Operating room Box Diagram.

Pr| E’\O') =1-Pr[ EVO] =1-85% =15%.

Column Total

iso, by independence, 13% =" 25%7a)

=>

a = 00%.

The complete Box Diagram is: Row Total

40

ee Pr| O|=40%

26.

(D)

60

(D)

A tree diagram would be nice, but would have infinitely many branches. The key is to figure out how many ways there can be a total of seven accidents in two weeks. There could be 0 accidents the first week and 7 accidents the second week. Or there could be | accident the first week and 6 accidents the second week, and so forth. Let (a,b) be the pair representing a accidents the first week and 4 accidents the second week.

Pr(exactly 7 accidents)

- Pr[(0,7) ]+ Pr[(1,6) ]+---+ Pr[(7,0) | =

ie gi

°

al 28

a

I cay)

prannerd ee ie Or i 2 o irst week

:

I yi prob of ; 7 accidents in

second week

Sealy

feeet

28

.

ie 2!

=

eel 99

=

Saran

64

°F: “ah

a oe

=

ee "y ——

’

et

:

ge
)' ies)

Pr(Z =i) = 5 fora

6

1 0}

18 38 (b) The first two throws must miss, and the third throw must hit.

(a)

PriS$ =1)=

ae Ps =.) -(#) (43) y

(c) p(s

5—]

=s)=( 22] (&) [Oty

= l,2, 50°".

>

(a) This is exactly the same probability distribution as Exercise 3-3. 18 (b) For example, shoot a 3-pointer until you make one, where p= —— 1s the 38

probability of success (making a 3-point shot). Di

52 @ SOLUTIONS TO EXERCISES - CHAPTER THREE

Section 3.2 Cumulative Probability Distribution

3-5 Value Of The Random Variable | Probability |Cumulative Probability

10/36 6/36

0

2

4

6

8

10

Sum of Dice

Cumulative Distribution Function

Sum of Dice

12

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

Bat

a) Pr = hyenG, (0.3) *-(0.7)* for #=0, 1.2.3, 4:

cpr |o0ost |0.0837 [03483 |0.7599 |1.0000| 3-8

Let H be the number of heads.

= Pri)

Pr[H >1] = 1—Pr[H =0] = 1-— = - * Then,

Pr sr

ST IRR

C3

|

t AY SAS 7 it) ole

is

F(t)=Pr(T = (.3)°

= 1.05.

+5-(.3)4 +675) +10-¢.3)? (75)? +10-(3)2 «(.7s)? +5-(.3)' -.7s)4 +€7sy°

60 40 1.00

66 ® SOLUTIONS TO EXERCISES - CHAPTER THREE

So the probability distribution for the random variable B is:

LB _| Probability | a a ee

Note: This is the probability generating function for a binomial random variable (wait for Section 4.2) with probability of success p=.7 and number of trials Peale),

3-48

_ (-.21)-(8)-G8))-(2)

(a) hg (s) = aaa]

77 aaa

_-

8

Geant

E[G]

oe. 3

eee

ee ORIG Ye errs

559 CVA aoe ars a Note: This is the probability generating function for a geometric random variable (wait for Section 4.3) with probability of success p=.8.

Section 3.8

Im

(a)

Chapter 3 Sample Examination

Pr(S 20] = 20-[) +2198) +22 [-B)+23-(-2) = 20.346.

(a) ELX] = .10(0) +.18(1) +.12(2) +.37(5) +.23(8) = 4.11. (b) ELX2] = .10(02) +.18(12) +.12(22) +.37(52) +.23(82) = 24.63. oy = ¥24.63-4.1122 = 2.78. (c) Q3=5. Let p denote the probability that the damage is $2,500. Solve 700 = E[ damage] =0-(.73) + 2500- p + 5000 - (.27 — p).

oN (a) E[damage |damage > 0] = 2500-52 + 5000-2 = 292.59.

2

.26

(b) E[damage? damage > 0] =2 500° an © damageldamage>0 a

(a) fy

te 5000- - oT

Ade ko:

= ELX] = 12.09) + 2(.32)+3(.42)+4(.13)+5(.04)

= 2.71.

The 50" percentile, median, is Y = 3. E[X?] = 17(.09) + 27(.32) + 37(.42) + 47(.13)+57(.04) Gy =8.23-2.717

(b)

£[Payment

= 8.23.

=.941,

at least 3 TVs].: = $75:2-2= alee e025 oe GTeS ines iy 59 eS)

Nn

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 71

LZ

(aye

elses tel 4el42 15.91 =
0] - 10-4100: = $40. (Cc) CVy

=

12 876

= 40.544%.

Ty

72

155

SOLUTIONS TO EXERCISES - CHAPTER THREE

SEIR SSE A) = SCS

202 2) ane

Gi eo, (48)

You might earn a raise if you informed your boss that

hp(s) = (.8+.2s)3

83 +3-.87 -.25+3-.8-(.25)? +(.2s) = .512+.384s +.09657 + .008s7

So the discrete random variable R has probability distribution

eS

Eee

CC 16.

eee eo

ee ee

Let x be the common difference between consecutive probabilities, po, p}.---, Ds From (1) po is the largest, with the remainder of the probabilities arithmetically decreasing in size. From (i) and (11) together we have, Mi

PossPe

CPE

Pe:

Then,

Py =(ParX;

and imeenetalsy First,

ase

7, =

Dia ps =

Pp = Pix

Dy iis Po

s,s

+ (Po

= (pox)— x — po

OL ee

—X) +++: + (po P|

Second, from (iii),

2

= potp

2x,

—5x)

=

6 po

—15x.

Ds

= pot (po-x)

= 2po-X.

Solve simultaneously the two equations, 6p9 —15x = 1 and 2pp —x

5

Po aT

: = 3 to get,

l

and x svat Then,

Pr| #Claims > 3 | =

pat ps

=(Po —4x) +(po —4y) = nee

fet

(C)

CHAPTER 4: SOME DISCRETE DISTRIBUTIONS Section 4.1 The Discrete Uniform Distribution

4-]

(b) To prove i

= 14+24+3+--4n

n(n+l) 5

=

i=]

by mathematical induction, we need to 1

I. Verify that the statement is true for n =1.

Me = [=

=vk

i=l II. Assume that the statement is true for n =k. ew

tiie: eyseal

III. Show that the statement is true for

[+2+34ek

+ (ke

= 9)

n=k +1.

(en

= oe.

(c) The statement is clearly true for the case n=1.

Assume that the statement is true for

Pepe ay

n=k.

k(k+1)(2k+1) 6

iy aay

Then,

+ (k+l)?

(k+1)| k(2k-+1) + 6(k+1) | 6

6 n=k +1.

showing the statement is true for (d)

The statement is clearly true for

N =0.

;

ataxt+ax* +ace+---+ax Rote ay

cee

6

Assume true for k+l ad—x*™")

l-x

+

l-x

a(l—x***) l-x

showing the statement is true for

ee

1104

i = Il

+

s|-Sei —

+

s|-— Saw —

We)

s|-

a

+

Se

to a ee) +

SS)

ars

+

N=k +1.

oo

“fe

=

S

se! Tal

—

———

\|

n-(nti) wn

9

N =k.

K+ ax*™(1-x)

Shue) 9

Then,

74

4-3

SOLUTIONS TO EXERCISES - CHAPTER 4

(b) The total accumulation of rice is 1+2+27+---+2°

= 2° —1 (geometric

series), which is a bunch more than one billion grains of rice. .

4-4

2

n?(n+1)*

We make use of the algebraic power sum formula | es =—_———.. | k=l es aes

~ 1 EV Xe | =o) Uke Ea 2 n

n?(n+l)* ee =

n(n+1)? See n 4

Then,

;

One may expand

E|(X-p)3| = E|X3-3uxX? +3n2X -13|

(APG RVI BC APREIILIA eye

II

= E[X?)-3uE[X7]+3u3 -pwe = ELX?]-3MELX2]4+2° This can be evaluated using the formulas, errr

vite

a

FLY?)

a

ee

and 3 (AO Ga i

n(n+l);

After a little algebra the result is seen to be zero. Or, one may simply observe that the values of X — w (and hence (Y—y)*) that are negative exactly cancel out the values of X — wu (and hence (Y¥—y)*) that are positive. Thus, the expected value, by this symmetry, must equal zero.

Discrete uniform distribution on {1,2,3,4,5,6'.

(b) EX] =

= 3.5,

(c) There are 6 data points, the median is

(d) oy =

67 -1

12

3+4 =3 2

Lo)

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

4-7

(a) Pr[(X =69) U(X =70)U---U(X =85)] = a (b) E[X]=

100 +1

2

= ehh,

(c) There is no mode. All 100 outcomes are equally likely. ip eae 100? -1

4-8

(a) Pr[Y 25] =5] = 1-+ = z. (yFIZ pe BI2 eX] = 2° RIX Ie 224) P10) (c) Var{Z] = Var[2-X] = 2? -Var[X] = +

5] = 26.6.

oz =5.164.

4-10

+45] = 5E[X]+45 = 5(3)+45 = 60.

-

xk b>) = Pr|A>= leer | = Pian

Il.

EY] = Ela-X +6] = a-E[X]+b

iy

Vary

|= Vvarfa-X +b) = a* -Var[| X].

—h

|

9)

# 75

76

Al?

SOLUTIONS TO EXERCISES - CHAPTER 4

a) PrLy is prime] ==, (b) PrLX >11]=30.

Section 4.2 Bernoulli Trials and the Binomial Distribution

4-13

E[M] = 0-(.1184)+1-(.3681) +2: (.3816)+3-(.1319) = 1.527. E[M2] = 02-(.1184)+1? -(.3681) +22 -(.3816) +32 -(.1319) = 3.0816.

o3, = Var[M] = E[M?]-(E[M]) = .749757. Ay WewePrMe=kh) ="35 Ce 8) 02) fork = 0,2 roi ELM |:=A2"(8) =.9.6 Var[M] = 12-(.8)(.2) = 1.92. Pr[M -(.6)? -(.4)'+(.6)? = 64.8%.

Sy ueaa

aera

Pr[2 < X]=1-Pr[X =0]-Pr[X =1]

=1-3C,-(.9)!-(.1)? — 3p -(.9)° -(.1)? = 97.2%.

E[X]=3-(.9)=2.7 and Var[X]=3:-(.9):(.1) =.27.

PROBABILITY AND STATISTICS WITH APPLICATIONS:

4-19

A PROBLEM SOLVING TEXT

® 77

A twenty question multiple-choice particle physics examination written in the language of Lower Lausatian, each question with four answer choices.

PrX=2] = 20C2-(.25) -(.75) TEP ET Ny 00-025) = 5 Pr[X =n-p-(l-p) = n- p-g. 4-26

Let N be the random variable that denotes the number of employees that achieve high performance. NV has binomial distribution with parameters n= 20 and p=2%. The first few rows of the probability distribution are

E24

Biles

ad 7 Pe

as

Prv=n) |6076 |2725 |0528 |. | Fo) [0675 [9401 |9929 |.

66.76% of the time, no employee would earn high performance, and C could be set at anything. Looking at the cumulative probabilities, 2 or fewer employees will earn high performance 99.29% of the time, so the probability of exceeding two awards for high performance is less than 1%. Thus, we can set C=120/2=60. (D)

4-27

The only time a penalty occurs is if all 21 ticket holders show up.

Efrevenue] =

21-50

—-100-

(.98)*!

——

= 1050-—65.43 = 984.57.

(EB)

—

revenue from sale of 21 tickets

probability all 21 people show

Section 4.3 The Geometric Distribution

4-28

Pr[D 0.2

.

=

2 0.2

a

3 0.2

P Binomial,

i

~ n= 10,

4 0.2

5 0.2

[= =

eee ead p=.3

3000

2000

-

4). 1000

0 0000

|

El

i)

a

1

BA

3

a

4

5

=

6

—

$3

—

=|

9 ;—

Prob 0.028 0.121 0233 0.266 0.200 0.102 0.036 0.009 0.001 6.000 0.000

ee yr te Hypergeometric

4-76 (c) 0.3000

0.2000

0 L000

OOOO

Prob

Ul

i

1

0.2166

04165

Z

OT

3

rn

0.0793

0.0086

a

5

0.0004

@ 89

90 # SOLUTIONS TO EXERCISES -CHAPTER 4°

4-76 (d)

|

|

ares

Poisson, Mean = 3 |os t

_

| |0.2000 |

;

| 0.1000 - ll 1Ih

» §.0090 1

i a

a

5 |

4-76 (c)

9

10

Prob0.049 0.149 0224 0m 0.168i 100 00.050 9.021 0.008 9.002 8.000

-

-

; Geometric, p=.25

7

0.3000

0.2000

0.19000

0.0000

-

| |

—

ii Hla

a

| ra)

i

4-76 (f)

TET

TE URGRS SIT

10

0.033 0.025 0.018 0014

or

Nesaive Binaitia r= 3 and p=.5 0 3000)

0) 2000)

0 1000

0 0000

I

9

10

Prob 0.125 0.187 0.187/0156 0.117 0.082 0.054 0.035 0.022 0.013 0.008

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

Random Variable [a RS irromia] Sng | pe yee

4-77

aa Pp

Negative Binomial

4-78

91

(a) E{coins}]=5)

© —_~

—

10gia 77-100

(b) on

mea 88.5.

(c)

Res =5 (2 ae12 = 1714, (b) ELX]

o arty=rounrant? «(2)2)2) —6X2+3] = 5-(3.933)-—6-(1.714)+3 = 12.378. E[5X

94

Di

SOLUTIONS TO EXERCISES - CHAPTER 4

(a) A=2 > = 4 sneezes per 2 hour period

(b) o=V4 =2 Ce

(d) 1-e*[1+4+8] = .762.

Fl=OX 213+281 =~ 6E[ X72] 413E[X] +28 - — 6{50-(.96)-(.04) + 487} +13-48 +28 = —13,183.52

6.

E[(3X +4\(7-2X)]

7.

(a) Pr(2 injuries next month) =

eo (3).

7

= .0747.

ee 51Go) (b) Pr(5 injuries next year) =

==a

|N, Number of Claims [0 [7a]

23

, pm

, 2

[om [SJ

, 3

(4) | 2

\|

-10 Nine

WE

/

mf

3

l | ] Pot (+) Pot (+) Po + Gg Porte a

- poyis(2)+(4) vf = rae

This implies that po =.8.

0.

Pr(N >1) = 1—.80-—.16 = .04 (A)

Let A be the number of people completing the study from one group of 10. Then 4 is binomial with n equal to 10 and p equal to .2. Then,

Pr| A >9] = 8194 19C, -.89 -.2! =.376. Let B be the number of groups in which at least 9 people complete the study. Then B is binomial with 1 equal to 2 and p equal to .376. Then, Pr{B =1] = (2)6376)\1=.376) = 469 (B)

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

10.

# 95

The scenarios where there are at least two more high risk than low risk include:

\hhhh, hhhm, hhhl, hhmm\. Nae 2) ale

n-p=8 Prex

Wer:

Cy (2) 03) £1, (25)

and n-p-q

10)

=

= 4.8 implies

20 Cio -(.4)!° =O)ue

=e

g=.6,

aCe (2)"23)2 = 0488 (D) p=.4,

and n=20.

Ly.

Let X be the number of low-risk drivers with no accidents in a given month and let - Y be the number of high-risk drivers with no accidents in a given month. Then X and Y are binomial random variables with,

E{_X ]=400-0.9 = 360 and E[Y]=600-0.8 = 480 The total number of bonuses paid over 12 months 1s, S = (X,4+:--+X12)+(% +--+ H2),

5S. E[5S] = 5(12-360+12-480)

and the total bonus payment for the year is

= 50,400

(B)

a

oe en

aca ee ‘

“oi

hp

)

7 tetas

Saray

shikai et Sard

Fe

hw

'

2%,

oe 4

|

neae Si seen Gieg eeca ‘

CHAPTER 5: CALCULUS, PROBABILITY, AND CONTINUOUS DISTRIBUTIONS

Section 5.2

5-1

Density Functions

(a) F(x) has the properties of a cumulative distribution function.

G@)

0

itpex MOQ)

P Mx() == Oe

ElX]=Myo)=£3 a by

(1-q-e')

Dp

legal raheemlee) 2

Pp

(ere)

1G (oe ee P?-4-2p-q-P-(-4} ae ee ee_g(p+2¢) Re »,_ eee Pp

Vari x)=

P 2

‘(p+ 42+ D2 isc

+2

ie

5-54

p

usto=| ie

1-ge'

Me

=Io( M0) =o

P

ayes l-qe

(0

Teter

}=r

P J=rincey rina-ge' ; —qe'

E[X]="@=—t.

}

meo=r (oe!)

5-55

Pp

euee

ke en

;

Dp

Pp

This is the MGF for a Negative Binomial random variable with r=4 and p=.3.

(b) Pr{V =2]= 5C3-(.7)° -(.3)4 =.0397.

5-56

0

Var[X]=h"(0)= Petre _

(a) My(t) een

a)

(b) My(t)=Mrx43(t) =e* ae

= ee" +31-6 |

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 111

oy

5-58

Yea

h(t)=In(M x(t)) = n(eHe) ale -1).

ELX]=h'(0) =.

Var[X]=h"(0) =A.

If Mz(t)= eles then the random variable has Poisson distribution with A =3.7.

The mode occurs at z =3.

Section 5.7 Chapter 5 Sample Examination

ies

(a)

ae fae 1=[

or = x

== [fax and Se

Solving for a and 5, we find the constants are

(ob)

ieee cee a=3.6 and

b=-2.4

Pr(X

m= S. and the mode occurs at x =0.

This is an example of Beta distribution. (a)

Tra= 12;

(b)

Pr X>.80)= [sgh2x(x-l)? jie aiDae

(c)

l l Var[X] = ih (x? )12x(x-1)? 2 dx -{[, (12x01? 2 as|; = .2-(.4)? 7 = .04.

(d)

The mode of the random variable X occurs at the maximum value of f(x).

f(x) = 12x(2)(x-1) +12(x-1)? = 12(x-1)[2x+x-1]. The critical points are

; ; | x =1 (not in the domain of f(x) ) and x= =

The mode is x =>.

(Awe

PA

eS) eas)

(b)

The mode is x=0.

(c))

ix =:

(d)

1-—e-*

(6)

sox =I)

= le?

= .80 implies xg9

= 551,

= —In(.2) = In(5) = 1.60944.

114 ®

SOLUTIONS TO EXERCISES - CHAPTER 5

12.

(a)

EfDamage] = (.92)-0+(.08)-1750 = 140.

(b)

ETE E[Ins. Pmt.] = (.92)-0+ (08)}say 0 —-

1000

1

3000

i

]

=10 0)- ee) as meas

.08| 800 | = 64.

13. (a)

My() = Ele] = [oe% Sede Ctds= = 5 [e :

14.

er

X=

=

ets]

=

fess

=

5—f

rcs.

: aay (ye 15 G20)

(b)

Ae (ey OnBr

(a)

Graphing the CDF allows you to see that X is a mixed distribution. The

peer

point masses are PrlX¥ =2]=.3 and

PrX =3] =.2.

(2) dv+3-(2) = oie24543See = (b) ELX] = 2-(3)+ fv ct Miners cers

5

J (x)

1S;

If My (t) a — 4e

then W has geometric distribution with p=.6.

(a) ow = Pe ) 16.

(b) — PrfW =3]=(.4) (6).

Since the man purchases the insurance at age 40, we are given that he has survived to age 40. E{Payment] = 5,000-Pr(40 An = 7 =

vea5

148 @ SOLUTIONS TO EXERCISES - CHAPTER 7

(c)

eee We have to find the marginal density function for Y, fy(v)= [= dx =—In(y), foreU=y =< i So

fy (1/4) e= In(4).

Sa (1s) al) F(i/4) ~ taf)” 4) - =f fx{v1¥=

Then,

i pesekal a iste ees) Ne ”

Pe x>Hy=t atin hoe

(d) ~~E(Net Loss] = ELY—Y]= i,[o--y)t PGi {de -+. (e)

Since Y given X =3/4 is uniform on ne/4],

elyix=3, _ 3/440 _ 3 2 (f)

7-43

8

“Off with her head!”

Fy (y) = [fy dx = 4y—3y’, for O< y5 dydx= (E x de==e

Cov(X,Y) = ELXY]-E[X]-ElY] = 24-63 =6.

(©)

24..

158 @ SOLUTIONS TO EXERCISES - CHAPTER 7

10.

Let O denote the number of tornadoes in county Q. Let P denote the number of tornadoes in county P. First, Pr[P=0] = .12+.06+.05+.02 = .25. Then,

jy Ve

ee Se

ee

it

Pry 22)

eS

09, 32 SS (.88) =. 980.

(D)

6 i hk (l-x-y) dydx

_

2 (l-x-y)?

= 6[/ere

l-x

x dx

_ 3 [ex de = - (ex) Bri 12.

Let let

8a= 488

G~ Exp(A =6) denote the random wait time for a good driver to file a claim and B ~ Exp(A =3) denote the random wait time for a bad driver to file a claim.

PG 3=W

+3,

we have

Ee Yi3) e735. o-and VarYY > 3) Var[W +3] Var[W]

= 5” = 25,

(A)

cae

atn

aif Sao

fr9rn Taw iy

m Ose

if

CHAPTER

8: A PROBABILITY POTPOURRI

Section 8-1 The Distribution of a Transformed Random Variable

8-1

For l = .8991. (E)

2 se fetta eT

8.2 The Moment-Gencrating Function Method n

=AD

=

n

y

i=l

|

Qa;

=

|

Q, +Q2+---+Qy

_

ot

M(t) = [[“« (t) = I Je

2

och ;

|)

j=

1

8-43

.

= ev

:

This is the MGF of a normally distributed random variable with

mean

is the probability

n

n

f=

i=l

"1; and variance }'o? .

Y"

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

# 179

8.3. Covariance Formulas

8-44 CovfU,V] = Cov[2X-Y,-X43Y] = E|(2X-Y)\-X+3Y)]- £[2X-Y]- E[-X+3Y] = E[-2X?+7XY-3Y?]-(2E[X]- ElY])-(-E[X]+32[¥)) = —2E[X?]+ 7ELXY]—3E[¥2]+ 2(ELX]) —7ELX]BLY] +3(E(Y]) = -2(ELX?]-(ELX])?) +7(ELXY]-£LX]- ELY]) -3(21Y?]-(E1Y))?) —2Var|X |+ 7Cov_X, Y]—3Var[Y]

—2(2)-3(3)+7(-1)

8-45 Cov[2X -4Y,X+7Y]

= —20.

2Var{ X]—28Var[Y]+10Cov(X,Y) = 2(5)—28(2)+10(-1) = —56.

8.4 The Conditioning Formulas 8-46

Using the conditioning formulas directly yields,

E(B] = Eg {E[B|G]} = .4(50) +.3(70) +.2(65) +.1(24) = 56.4. Var[B] = Eg \Var[B| G]} +Varg {EB |G]}

are ae }:2(652) +.1(0) +|.4(50-56.4)? +.3(70-56.4)? +.2(65-56.4)? +.1(24-56.4)?| Here is the complete calculation in tabular form, modeled on Examples 8.4-1 and 8.4-2, where_X is the sporting event attended and Y is the quantity of beer consumed: Pr{X] | E[X|¥] | Varly|X1 | ELEY X]] | £[veri¥ |X1] | Eley xP] 6.4

20

Cubs

aa

or

=

—

1470

845

845

E(Y]= a X]]=56.4

E[Y|X] = ELey | XP |- ely) = 3372.6—(56.4) = 191.64, Var E|Var[Y |X]]}=1101.4, and

Var[Y] = E[ Var[¥| X]]+ Var| E[Y | X]] = 191.64+1101.4 = 1293.04 =

Oy =55.96

1000

250

180 ® SOLUTIONS TO EXERCISES - CHAPTER 8

8-47 (a) ELX] =Ep{E[x|6]} = Ey 2 2-5 mifhion: (b) Var[X] = Eg {Var X |6]} + Varg {ELX |0} Fo

62 0 D |Yar {8)

I

3h h+ qo

To Bet?

I Seer 2252 nis: Soi Fa (million)° =>

=

Prev > he) =

poe

18}

Me oy = 3.323 million

0 fig Ot ta Pees 6

Pree ee joe

d0+ ,

x -e-5d@ = 5742. 5

8-48 (a) E[S] = Ey {£[S|N} = E[X,+X2+---+Xy]

EL X,]+ E[X2]+---+ E[Xy] = N- py = un: Mx.

(b) Var[S] = Ey Var{S |N}}+ Vary {E[S| N]} = Ey{N -Var[X]} + Vary {N-ELX]} Var[X]-ELN]+(ELX]) -E(N] = E[X?]-E[N]. Note:

8-49

See also property (2) for random sums.

If we let T denote the total losses, then

Sern l

12,500

Pp

E(T |N]= ux N =12,500N

15,0007 /12

and Var[T|N]=o2N=

15 ,0007

12

E(T]= E[E(T |N)]= EluyN]= wx ELN]=12,500p Var[T] = En \Var{T |N}} +Vary {E[T |N]} 2

“

é)

7

=Var{T |=Ey \o} N}+Vary {uN} =03 pt ui pq =

Ss= +12,500? - pg

lis

)2

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

@ 181

8-50 In the case of partial damage the expected payment in thousands is given by, by parts with w=x-1

and dv=e-*/?dx reoOF

J, @-1)-[.5003-e~*?dr] = .5003 f“ete ak 15

5003[-2e"(x+1)]

II

15

= 1.2049

]

Exe

ReymeeelDI 1.2049 pele

E| Payment] = E [£(Payment |Damage) |

= .94-0+.04-1.2049+.02-14 8-51

= .3282 thousand

This is an exercise in detail. As an example, we calculate

(B)

Pr(S =4). Think about all

of the ways that the sum could equal four:

Pr(S =4) = Pr(X¥ =0,Y =0,Z =4) + Pr(0,3,1) +Pr(1,1,2) —.

FH

denoted

Pr(0,0,4)

+ Pr(1,3,0) + Pr(2,0,2) + Pr(2,1,1)

= (.4)(.25)(.55) + C4)C4)GC15) + (3)¢35)(.2)4+ (3). 491) + (.3)(.25)(.2) + .3)(.35)C15) = .14275 S=X+4+Y+Z

0.0365 0.07025 0.0965 | 0.14275 0.20125 0. 141 |

N]rRe |] BI] NH} DA] CO}N] OO}

EUS] = 0-.01+---+9-.066 = 5.2

0.16975 0.066 0.066

182

SOLUTIONS TO EXERCISES - CHAPTER 8

8-52

EY] =

y-2-(l-y) dy = 7 Let N be the number of claims, a binomial random {,

variable with n=32 and p=1/6.

= Ven

Let X,...,Yy be the policies with claims and let

Ae

[S] = E[ES|N]| =wy -sy 8-53

Let S=¥,+¥%)+---+Yy

8-54

ae 68

16

denote the random sum.

RN oD= 32, p=). ee N ~ Binomial(n Val S\=00

Leal mite

sy =+ and o; ==.

_16 and oy Fa NIELS 37.1 eS We Prarie

37 a =32 2h a:

ale Opal On iy + Lino far = LORE E eae

The random variable T|.X ~ Unif[X,2X]. Soper

fe a)

ee 0x

ef

0

otherwise

2x-3 yp Pr(T (i) = — : ——_ 5 dxeS

id = 64

x>15 ere

Sal.

Let N be the number of people hospitalized. The unconditional probabilities for N are,

Pr[L »

Leap e9 a

= In(.25-.50-.40-.80-.65)

= 137

2 ee)

. Thus,

(B)

Let gq be the probability of a$500 loss. nent pa giao

2 peel — gs | >

pss.

Since there are 3 outcomes on each ofthe 3 trials, the probability of observing the

particular outcomes $0, $0, and $1000, using the multinomial formula, is, 3

2

Since p
0, oy

n

and hence J? is consistent. Note: There is a minor complication here since T? is only asymptotically unbiased.

This is because T* = 415 2 where S? is the unbiased estimator of

o*. Thus, the argument showing that the variance of T” converging to zero implies consistency requires a slight technical modification.

In f(x) =In3 + InO+2Inx-0x3 => LACES 00

See ge

The CRLB is given by —

oO)

:

:

na]ome

Un f2)] =—

0

= —

: —

=

Ce

is

git

(E)

06?

et

ae

nel

U.

s

1

ee

In Example 11.3-1 we showed that Var[@,] =

n(n+2)

Since this converges to zero as n goes to infinity, the estimator is consistent.

11-29

Since B is binomial, E[B]=np and Var[B] = np(l— p). Let p= ity n Ronn Joes | p(l-p) Then, Z| p|=p and Var[p] = ——-np(l-p) = ———-. (a

ai

Since this converges to zero as n goes to infinity, the estimator is consistent.

11-30

In f(x) = In A — Ax = The CRLB

is given by

A

Vise

sul

A)2

hy:

a ete ee aA2

ng|Slnfe:2) |

=

RA

ary i )

So

n

(E)

*

97,

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 217

Section 11.4 Hypothesis Testing Theory 11-31

(a)

We use a standard normal z test based on the sample proportion p= in where X is the number of made shots in 100 tries. The rejection region is

A — 80 z -80-.20 100

X¥73| p=.5] = Pr) Z>

11-32

esas V100-.5-.5

| mee

Use a right one-tailed rejection region of the form X > 4, where,

05 = Pr x2 4|u=10|)= Prizes | >

A=" 1.645.

Vl/n

lin

06 = LF k , where k is the smallest integer for which, Pr[N >k] < .10.

From the table, we have k =3, with

Pr[N >k] = 1-.92 = .08 < .10. Then, the power equals,

1-8 = Pr[N>3|A=2] = 1-Pr[N =0,1,2|/ =2] 50

=

51

e722

52

—

7

= ]—5e-

= 325

(B)

218 ® SOLUTIONS TO EXERCISES - CHAPTER | |

11-34

11-35

The significance level is the probability of rejecting Hy when Ho false. II. is true, and III. is false. (B)

is true, so I. is

likelihood ratio critical region is of the form, | and so, the L(A) =An"eAMte+"+%) e(nitx E+E) EX, ) L(1)

LQ)

e(™ +X

|” el

Ftp

Q*etarmetm) 4++-+Xp_)

=< 2° k =e k'

on

Syy =—112.695 (negative since the correlation is negative).

be a Pond 0 AY ORY =1137 oY Sao

Then,

tt

oor,

XX

For X¥=6,Y 11-60

=d+BX =11.37-.69-6=7.251.

Using the given data for_X and Y, we have,

N =5,X =3,Y =13,Syy =10,Syy =34,Syy =17. Then,

R? ee

eee(or)5.

(D)

Sixx Sy

Note:

Since the values of Y, are also given, one could calculate ESS and use

formula (8): ESS = Syy -(1—-R?).

11-61

Syy =10,Sxy =195=> berSxy =) 719.5, and Sixx

@=¥ — BX =45-(19.5)3) =-13.5 > Y =@+BX =-13.5419.5-1.2=9.9. (B) 11-62

N=d,X s45.Y =1.8286- Sp =700,Syy

0804s Sap = nS

p= 2X = 0336 and @ =Y—-BX =.3179 > ¥ =@+B-75 =2.8357. (A) 11-63

N=12,X =12,¥ =145.1667,S yy =572, Syy =59,793.6667,S xy =5792=> ie =10.13 @=Y-BX =23.66 = Y =23.66+10.13.X (E)

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

227

11-64 We have InY =Ina+ BX , and,

Ce

Rea ee

ree

ase NS|B 9119.01 7.030" |Om 49A 221,098)

Prowisp is] [3523 |55|248.28|106.23 N=5,X =3,In¥ = 7.0463,Sxv=10,S(inyynv) =-0294,Sxiny =-5314 =>

B =23+4 =.0531 and In@ = 7.0463-.0531-3 = 6.8869 =>

InY = 6.8869+.0531-6 = 7.2057 =>. Y = e797 = 13475. (D)

39, OT

11-65

11-66

=g _ 23 ee 7ee _ BF ers s

OT RC

N=5,X =7.1,Y =1.06,Syy =.2,Syy =.332,Sxy = 18> (a)

p= Sa =.9 and @=Y - BX =-5.33

(b)

Y=-5.33+.9-7=.97

(c)

2_ Ra

(d)

ESS =Syy -(I-R?)=.17

(ec)

TSS = ESS+RSS

Sey ae

_= 488

=>

RSS = TSS—ESS

= .332-—.17 = 162

11-67

wehave

Y = a@+ PX , so that the regression line passes through

Since

@=Y —BX

Dah

Minit eH G@+2f. Also, 3.3 = a+3B.

Eliminating @ between these two equations gives,

B=3.3-Y.

228 @ SOLUTIONS TO EXERCISES - CHAPTER | |

Next, note that Y = fo"

=

"w = 5Y —12° Then,

Sxy = (0)(5)+()G.5) +(2)3) +3w+t(4)(5)- S)(2)Y =129

5 3(57 -12)_-10Y = 5Y —65,

Now, £= 22158 fete OO Combine this with the previous expression, Sixx 10

i= ieee aver

aS

eee,

Sy

282 ey 19 ae

This implies Y = 30 and i Sef

11-68

Weused Excel to generate the following chart.

Actuary Income

§ 60000 ”

o 50000 E[e)

S 40000

y = 2884.9x + 43759 R?ae= 0.9119

Se

#® 30000 0

1

ie

3

4

)

Number of Examinations Passed

(a)

Notice that the regression equation y = 2884.9x + 43,759 should be interpreted

as follows: The intercept of 43,759 should be thought of as a base salary. (b)

The slope 2,884.9 implies for each examination passed, companies compensate their employees an additional $2885. So if Hilary had passed five examinations while in school, we would predict that her starting salary would be near y = 2884.9(5)+ 43759 = $58,184.

Note: The accuracy of the regression equation is measured by R-. However, ifthe data is not chosen randomly or disobeys our assumptions, then our work and associated statistics are meaningless.

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 229

Procedure to Use Excel®

11-69

li

Enter the data in two adjacent columns, x first, then y.

2.

Highlight the data and make an XY scatter plot. You should label the axes and the chart.

3.

Left-click on one of the diamonds then right-click to add a trend-line. Choose a linear trend-line for simple linear regression. If a different model fits the data more closely, then another model may be preferable. Such a discussion is beyond the scope of this text.

4.

Click on the trend-line to format it. Under options, add the R-squared value and the equation.

(a) Salary and Weight 400

350

y = 9.7297x - 189.05 R? = 0.997

Ena

| |

2

|

2g

| |

=

55 Salary

(b)

The correlation is ./.997

cenit

=.998 (positive, since the slope is positive).

_39 99. ¥ =200 then ¥ = 220+187:09 Oey 5

230

SOLUTIONS TO EXERCISES - CHAPTER | |

11-70 (a) Children and Income

y = -2.9173x + 24.898

20 4

R? = 0.8329

10 | (1000) Income Disposable

af >

0) 0

2

4

6

8

10

Number of Children

(b)

The correlation is —/.8329 =-—.9126 (negative, since the slope is negative).

OMT at 4hen

9 0173

04 RoR

FI29F

ee

11-71 ACT and Grade in Stat

400

.

2

2 90 8 S a

*

y = -2.5691x + 140.58 R2=0.9723

80

& ®

PRS

6 | £ 60 [tS

|

a

50 |

40

15

20

25 ACT Score

Correlation is —4/.9723 =—.986.

30

35

40

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT

Section 11.8 Simple Linear Regression: Estimation of Parameters 9)

11-72

Given Var[o7]=4.

From formula (16), Var[ B= ca . From the data, DOG

Seat co Var p\== ==. GE

CE (n-2)

B

oi

gee

S yy

11-74

a0

Sie;

Syyv

ara

as

i

ge fe ae

ar

Ve GsYe A |

Sixx

Formula (26) gives,

Pe eeSS Te (N=D)RSS= (N= 2)RSS Ts , ESS /(N—2) ESS TSS. ESS” From formula (15), TSS = RSS+ESS. Also, R? = = $0 ESS TSS : 1 Th es RSS ESS ype 5 ESS! TSS’ TSS ame ns Paks iy Le He oo

S89)

TSS

ESS

2

1— R?

From the data, N =20,¥ =5,Y =6,Syy =1600,Syy = 6400, Syy =1920=>

peek,

SAC

Saeed 7 asm 100086400 (18)(.36) Mares CV=2 RSS Ber SS BB) i S516

0 105.

® 231

232 SOLUTIONS TO EXERCISES - CHAPTER | | a elit i el at de et Ce

;

1-75

pe Ory ee 1okU = = i Sxx 1600

eS

no

eleee

=1.2 and 6 = Y-BX = 6-1.2-5 =

Then, Y =1.2X. For X¥ =10,Y =12.

) 6 = S?= ESS. z ee ee =409 = (1-.36)(6400 ESS = (1-R?)JTSS tos (18) = 1.734. (a)

Using formula (23), Y% + ts/2(N—-2) ; ; rv is, 12 ++1.734 interval

|2, the confidence

1 (10-5) 2 pee 1600 Jexr56)= (5.3,18.7).

(b) Using formula (24), Y + ts;2(N-2)

interval is, 12+1.734

11-76

i eH N Sixx

fe

1 +—+ eB) S* , the confidence

OSE |ae SO)

We use the relationship, (1, N—2)= ies — R?

,

= (—15,39),

with

Nos and ho 67 36 hen es) -= at 212.942 From the F table ae)

Fos (1,23) =4.28 , and /'¢, (1, 23) = 7.88. rit reject in each instance.

11-77

ESS =7SS —RSS =43—37 =6.. Then;

F(,N-2)

=

RSS.

—lg~ = aitSA Al oes aay Rl

41.83 =11.91.

(n-2)

Section 11.9 Chapter 11 Sample Examination

ie

Maximum Likelihood:

L(@) = O°(x-+-x5)?!

eGae.

=>

InZ(@) = 5In@+(O-1)In(x:--xs)

ae rae

=> {InL(O)) = F+ini%s)=0

Ale

>

Method of Moments:

ELX] =

Gives §=5

=1.07, Then, R—S

=

Ox® dx =

R

i

osae

in(21-43-56-.67.72)

aa and X = .27.

(A)

=13

>|

=.518. Equating these

Nn

PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 233

Ay: fo(x) = 1/10 for 0 Fos(12,11)=2.79. 05 = Pr

i2 -

==

ave

i

Se

o

FQ IN) 2 a IRB O.5o)

Oe PL =

OK SZ,

22.79 (E)

Then,

234 @ SOLUTIONS TO EXERCISES - CHAPTER | |

5:

Let X ~ N(20,27) be the side of the square.

£ Z Then, Y ~ N [20,22 ;

The estimate for the area is X? and E[X?] = Var[X]+E[XP?

=

This will overestimate the true area by .8 meter. (D)

N

=7,W =1.86,Y =6,Sww =18.6,Syy =138,Swy =50

p= ae =2.65 and @ = Y-BW

=108

=>

=> Y = 1.08+265/X.

AX

ab

Let_Xbe lognormal, with XY =e”, where W ~ N(,07).

From the properties of the

lognormal distribution in Section 6.3.4 we have, E[X]=e4*(/)>"

ELX?]=e2#+2"

Then, InELX] = io

o? = InE[X*]—2InE[X].

and InELX?] = 24+207.

Thus,

The method of moments estimator for o? is

In 4.4817 —21n1.8682 =.25 , and so the estimator for

8.

and

Since X is exponential, E[X]=6 and Var[X]=6?.

Pr[Type II error] = Pr[_¥